Please help me with these AP Chemistry homework problems and show me the steps how to find the solutions:

Liquid-Liquid Tirations: What is the concentration of the unkown solution for each of the following titrations?

#5) 25.36 mL of 0.178 M NaOH is required to titrate an 25.00 Ml of an unknown concentration HF acid solution?
NaOH(aq) + HF(aq)--> NaF (aq) + H2O(l)

#14) 45.32 mL of 0.0200 M KMnO4 is required for complete titration of an 7.00 ml sample of decomposed blood. What is the molarity of iron in the blood sample? (Note: The blood sample is treated to destroy the hemoglobin and then to reduce iron to iron (II) ion. These steps have no affect on your calculations.)
8H+(aq) + 5Fe2+(aq) +MnO4- --> 5Fe3+(aq) + Mn2+(aq) + 4H2)(l)

#19) 34.09 mL of 0.251 M KOH is required to titrate a 10.2951 g sample of acetic acid. The density of the solution had been prevously found to be 1.007 g/mL. What is the molarity of the acetic acid solution?
KOH(aq) + HC2H3O2(aq) --> KC2H3O2 (aq) + 2H2O(l)

1. Write the equation and balance it.

2. Calculate mols titrant added. mols = L x M
3. Convert mols titrant to mols unknown. You do this conversion with the coefficients in the balanced equation. Just the same as dimensional analysis.
4. Convert mols unknown to Molarity from mols = M x L

Sure, I can help you with these AP Chemistry problems. Let's go step by step to find the solutions:

Problem #5:

Given information:
- Volume of NaOH solution used: 25.36 mL
- Concentration of NaOH solution: 0.178 M
- Volume of HF acid solution: 25.00 mL

To find the concentration of the unknown HF acid solution, we can use the concept of stoichiometry. By looking at the balanced chemical equation:

NaOH(aq) + HF(aq) -> NaF(aq) + H2O(l)

We can see that the ratio of moles of NaOH to moles of HF is 1:1. This means that the moles of NaOH used will be equal to the moles of HF in the solution.

Step 1: Convert the volume of NaOH solution used to moles.
- Moles of NaOH = Volume x Concentration
- Moles of NaOH = 25.36 mL x 0.178 M

Step 2: Since the ratio of moles of NaOH to moles of HF is 1:1, the moles of HF will be equal to the moles of NaOH.

Step 3: Convert the moles of HF to concentration.
- Concentration of HF = Moles / Volume
- Concentration of HF = Moles / 25.00 mL

Now, let's plug in the values and calculate the concentration of HF.

Problem #14:

Given information:
- Volume of KMnO4 solution used: 45.32 mL
- Concentration of KMnO4 solution: 0.0200 M
- Volume of blood sample: 7.00 mL

To find the molarity of iron in the blood sample, we need to use stoichiometry again. Looking at the balanced chemical equation:

8H+(aq) + 5Fe2+(aq) + MnO4-(aq) -> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

We can see that 5 moles of Fe2+ react with 1 mole of KMnO4. Therefore, the moles of Fe2+ will be equal to the moles of KMnO4 used in the titration.

Step 1: Convert the volume of KMnO4 solution used to moles.
- Moles of KMnO4 = Volume x Concentration
- Moles of KMnO4 = 45.32 mL x 0.0200 M

Step 2: Since the ratio of moles of KMnO4 to moles of Fe2+ is 1:5, the moles of Fe2+ will be 5 times the moles of KMnO4.

Step 3: Convert the moles of Fe2+ to molarity.
- Molarity of Fe2+ = Moles / Volume
- Molarity of Fe2+ = Moles / 7.00 mL

Now, let's calculate the molarity of Fe2+ in the blood sample.

Problem #19:

Given information:
- Volume of KOH solution used: 34.09 mL
- Concentration of KOH solution: 0.251 M
- Mass of acetic acid: 10.2951 g
- Density of solution: 1.007 g/mL

To find the molarity of the acetic acid solution, we again use stoichiometry. Looking at the balanced chemical equation:

KOH(aq) + HC2H3O2(aq) -> KC2H3O2(aq) + H2O(l)

We can see that the ratio of moles of KOH to moles of acetic acid is 1:1.

Step 1: Convert the volume of KOH solution used to moles.
- Moles of KOH = Volume x Concentration
- Moles of KOH = 34.09 mL x 0.251 M

Step 2: Since the ratio of moles of KOH to moles of acetic acid is 1:1, the moles of acetic acid will be equal to the moles of KOH.

Step 3: Convert the moles of acetic acid to molarity.
- Molarity of acetic acid = Moles / Volume
- Molarity of acetic acid = Moles / (Mass / Density)

Now, let's plug in the values and calculate the molarity of the acetic acid solution.

I hope this helps you understand how to approach these liquid-liquid titration problems in AP Chemistry. Let me know if you have any further questions!